Question

Help please.
Does this series Converge/Diverge?
sum(1 to infinity) ((2n)!)/((n!)^2)

is (2n)! = 2!n! ?
I tried the Ratio test and it was CONVERGENT with the idea that (2n)! = 2!n! and thus, n! will cancel one of them in the denominator. With this, it was convergent, but when I tried wolfram, it was DIVERGENT by limit test (or Test for Divergence).

Thanks for helping.

Answer 1

\((2n)!=2!n!\) is not true, though.

Consider \(n=2\). Then you have
\[(2\cdot2)!=2!2!\\
4!=2\cdot2\\
24=4\]
(not true)

Answer 2

Oh right, thanks. hmm... I'll be right back, need to eat lunch

Answer 3

lim(n->infinity) abs( a_(n+1)/ a_(n) )

= lim(n->infinity) (((2(n+1) )! / ( (n+1)! )^2)) * (((n!)^2) / ((2n)!))

= lim(n->infinity) (2(n+1))/(n+1)^2
= 0 <1, which by Ratio Test, it will converge.

so...why does wolfram say it diverges?

Answer 4

my equation maker is messed up..sorry.

Answer 5

After my simplifications I end up with:

\[\frac{ 2(2n+1) }{ n+1 }\]

for a quotient, which turns out to be 4 as n approaches infinity

Answer 6

If \(a_n=2n\), then \(a_{n+1}=2(n+1)=2n+2\), not \(2n+1\). I think that's the mistake you're making.

Answer 7

@SpaceLimbus, interesting, I have a different fraction but I get the same limit...

Answer 8

An = 2n? where did that come from?

Answer 9

It's quite possible that I messed up at one step (-: But we both agree that it's divergent then, including the wolf from the post apparently.

Answer 10

you simplified the (2n)! / ((n!)^2)?
how?

Answer 11

\[(2n)! / (n!)^2\] <--- from this, how did you come up with 2n?

Answer 12

@franciscanmonk , IN GENERAL, if \(a_n=2n\), then the next term in the sequence, \(a_{n+1}\) is \(2(n+1)=2n+2\). I'm not saying this is the \(a_n\) for your problem. Sorry for the confusion.

Anyway, using the ratio test, you have the following limit:
\[\lim_{n\to\infty}\frac{(2n+2)!}{((n+1)!)^2}\cdot\frac{(n!)^2}{(2n)!}\]
Right?

Answer 13

\[(2(n+1))!=(2n+2)!=(2n+2)(2n+1)(2n)!\] which should simplify with your denominator, just keep on applying that to your problem as well as for ((n+1)!)^2

Answer 14

I see... Ok

Answer 15

Along with that, you have
\[((n+1)!)^2=\bigg((n+1)!~(n+1)!\bigg)=\bigg((n+1)n!~(n+1)n!\bigg)=(n+1)^2(n!)^2\]

Answer 16

\[\lim_{n\to\infty}\frac{(2n+2)(2n+1)(2n)!}{(n+1)^2~(n!)^2}\cdot\frac{(n!)^2}{(2n)!}\\
\lim_{n\to\infty}\frac{(2n+2)(2n+1)}{(n+1)^2}\\
\]

Answer 17

OH!!!! HAH!!!! There it is!! Thank you guys!! I love you both!!!

Answer 18

\[\frac{ (2n+2)(2n+1) }{ (n+1)^2 }=\frac{ 2(n+1)(2n+1) }{ (n+1)^2}=\frac{ 2(2n+1) }{ (n+1) }\] just to complete the algebra, I first overlooked that myself, so you have equal orders

Answer 19

You're welcome!

Answer 20

Ah, I see now. I just looked at the coefficients since the degrees of numer-/denominator were equal.

Answer 21

I can only give 1 medal, though... which one of you wants it?

yeah, the 2 remains,
and 2 > 1, which in ratio test is divergent

Answer 22

wait, that would be 4

Answer 23

give it to @SithsAndGiggles, in general if you want, give it to the first mentionworthy contributor that began to help you with your problem, I only joined in because I'm an insomniac with a fable for late night math sessions ^^

Answer 24

haha. Ok. Thank you again, both of you. :D